Question

I got this question at an interview and at the end was told there was a more efficient way to do this but have still not been able to figure it out. You are passing into a function an array of integers and an integer for size of array. In the array you have a lot of numbers, some that repeat for example 1,7,4,8,2,6,8,3,7,9,10. You want to take that array and return an array where all the repeated numbers are put at the end of the array so the above array would turn into 1,7,4,8,2,6,3,9,10,8,7. The numbers I used are not important and I could not use a buffer array. I was going to use a BST, but the order of the numbers must be maintained(except for the duplicate numbers). I could not figure out how to use a hash table so I ended up using a double for loop(n^2 horrible I know). How would I do this more efficiently using c++. Not looking for code, just an idea of how to do it better.

Answer 1

In what follows:

1. arr is the input array;
2. seen is a hash set of numbers already encountered;
3. l is the index where the next unique element will be placed;
4. r is the index of the next element to be considered.

Since you’re not looking for code, here is a pseudo-code solution (which happens to be valid Python):

arr = [1,7,4,8,2,6,8,3,7,9,10]
seen = set()
l = 0
r = 0
while True:
  # advance `r` to the next not-yet-seen number
  while r < len(arr) and arr[r] in seen:
    r += 1
  if r == len(arr): break
  # add the number to the set
  seen.add(arr[r])
  # swap arr[l] with arr[r]
  arr[l], arr[r] = arr[r], arr[l]
  # advance `l`
  l += 1
print arr

On your test case, this produces

[1, 7, 4, 8, 2, 6, 3, 9, 10, 8, 7]